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## SO(3) and SU(2)

In order to study the hydrogen atom, we’ll need to know something about the representation theory of the special orthogonal group $\text{SO}(3)$. This post consists of a few preliminaries along the way to doing this. I’ll be somewhat vague about a few things that 1) I don’t have much experience with, and 2) that would detract from the main narrative anyway.

Rotations

First some generalities. $\text{SO}(n)$ is the group of all rotations of $\mathbb{R}^n$ fixing the origin. Abstractly, we equip $\mathbb{R}^n$ with an inner product and an orientation, and then $\text{SO}(n)$ is the group preserving both these structures. As a subspace of $\mathbb{R}^{n^2}$ cut out by algebraic equations, $\text{SO}(n)$ inherits the structure of a smooth manifold which is compatible with its group structure, making it a Lie group.

By the spectral theorem, any rotation $g \in \text{SO}(n)$ has an orthonormal basis of (complex) eigenvectors. Since elements of $\text{SO}(n)$ preserve length, the corresponding eigenvalues must all lie on the unit circle $\mathbb{T} = \{ z \in \mathbb{C} : |z| = 1 \}$. If $v$ is an eigenvector of $g$ with eigenvalue $\lambda$, then since $g$ is a real matrix, the complex conjugate $\bar{v}$ is an eigenvector of $g$ with eigenvalue $\bar{\lambda}$. From here there are two cases:

1. If $\lambda$ is real, it equals $\pm 1$, and $v$ must be real.
2. If $\lambda$ is complex, $v + \bar{v}$ and $\frac{v - \bar{v}}{i}$ are real, and $g$ acts as a rotation by $\arg(\lambda)$ on the real subspace they span in $\mathbb{R}^n$.

Moreover, since $g \in \text{SO}(n)$ its determinant must equal $1$. Since complex eigenvalues come in conjugate pairs, it follows that the eigenvalue $-1$ must occur an even number of times, so the corresponding eigenvectors can be paired up into 2-dimensional subspaces of $\mathbb{R}^n$ on which $g$ acts as rotation by $\pi$. This gives a fairly concrete structure theorem for rotations: any rotation fixes a subspace of even codimension and acts as a direct sum of rotations in $\text{SO}(2)$ on the complement of this subspace. (This gives a first indication that the behavior of the group $\text{SO}(n)$ depends strongly on the parity of $n$.) By varying the angle of rotation in each part of this direct sum, it follows that $\text{SO}(n)$ is path-connected.

Specializing to $3$ dimensions, any non-identity rotation has a unique eigenvector of eigenvalue $1$, its axis of rotation, around which it rotates acting as an element of $\text{SO}(2)$. Concretely, every element of $\text{SO}(3)$ is conjugate to a matrix of the form

$\displaystyle \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \theta & - \sin \theta \\ 0 & \sin \theta & \cos \theta \end{array} \right].$

These are the maximal tori in $\text{SO}(3)$.

The universal cover

Suppose a quantum system (that is, a Hilbert space $X$ and a Hamiltonian $H : X \to X$) has a path-connected Lie group of symmetries $G$. The previous post may have misled you into thinking that this implies that $G$ has a unitary representation on $X$. Actually, it implies something more subtle. Instead of a unitary representation of $G$, what is physically relevant in quantum mechanics is that we have a projective unitary representation of $G$. Since the space of states of our quantum system is not really the unit vectors in $X$ but the unit vectors in $X$ modulo phase (the projective Hilbert space $\mathbb{P}(X)$), symmetries of our quantum system should really be symmetries of the projective Hilbert space. The group of all possible such symmetries, rather than the unitary group $\text{U}(X)$, is the projective unitary group $\text{PU}(X)$, and an action of $G$ on our system is then a morphism $G \to \text{PU}(X)$.

Any unitary representation $G \to \text{U}(X)$ gives rise to a projective unitary representation, but in general there will be extra representations which do not arise in this way. This has some curious consequences. Consider a smooth path $p : [0, 1] \to G$ such that $p(0) = p(1) = e$. This path describes a smoothly varying transformation of our quantum system; for example, if $G = \text{SO}(3)$, such a path might be describe a rotation about some axis by $2 \pi$. After applying this smoothly varying transformation, the fact that we end at the identity implies that we have the “same” quantum state as we did before. But this only means that the old state $\psi \in X$ and the new state $\psi' \in X$ differ by scalar multiplication; they don’t have to be the same unit vector in $X$.

(The projective representation of $\text{SO}(3)$ we will primarily be interested in is its representation on $L^2(\mathbb{R}^3)$, which comes from an ordinary representation. However, it is extremely important to understand the other projective representations of $\text{SO}(3)$, as they relate to the topic of spin, and spin will come into our final calculation of the sizes of electron orbitals.)

After making the physical assumption that the phase we gain after applying $p$ to our system only depends on the homotopy type of $p$, it follows that physically relevant projective representations of $G$ come from ordinary representations of its universal cover $\tilde{G}$. Concretely, this is the group of all pairs $(g, p)$ where $p$ is a homotopy class of paths $[0, 1] \to G$ such that $p(0) = e, p(1) = g$ (roughly speaking, all distinct ways to smoothly apply $g$), and multiplication is pointwise. The covering map $\tilde{G} \to G$ is the map that forgets the path $p$, and it is a morphism of topological groups with kernel the fundamental group $\pi_1(G)$ (based at the identity), which sits inside $\tilde{G}$ as a normal subgroup; that is, there is a short exact sequence

$\displaystyle 1 \to \pi_1(G) \to \tilde{G} \to G \to 1$.

Theorem: As a subgroup of $\tilde{G}$, the fundamental group $\pi_1(G)$ is discrete and central (in particular, it is abelian).

Proof. The fact that $\pi_1(G)$ is discrete follows from the fact that Lie groups, being smooth manifolds, are locally path-connected and semi-locally simply connected (hence small perturbations of a path in $\pi_1(G)$ are homotopic to it). It is also true generally that a discrete normal subgroup $I$ of a connected topological group $H$ is central. To see this, given $i \in I$, consider $C_i = \{ hih^{-1} : h \in H \}$. Since $C_i$ is the continuous image of a connected space, it is connected, but by normality it is a subspace of $I$, and by discreteness it must be $\{ i \}$.

(The fact that $\pi_1(G)$ is abelian also follows from the Eckmann-Hilton argument: there are two multiplications on $\pi_1(G)$, one being the usual fundamental group multiplication and the other being pointwise multiplication, and they respect each other, so they are identical. This is why $\pi_1(G)$ actually sits inside $\tilde{G}$ as a subgroup.)

In Lie theory, this theorem has the following important consequence: to classify Lie groups, it suffices to classify simply connected Lie groups, then to compute their centers, then to find all discrete subgroups of their centers. For our purposes, it tells us that to find projective representations of $\text{SO}(3)$ we should find its universal cover. The abstract name for this universal cover is the spin group $\text{Spin}(3)$, but in dimension $3$ there happens to be an exceptional isomorphism between $\text{Spin}(3)$ and another Lie group which we’ll exploit.

The fundamental group

Whatever the universal cover of $\text{SO}(3)$ is, it sits in the short exact sequence

$1 \to \pi_1(\text{SO}(3)) \to \text{Spin}(3) \to \text{SO}(3) \to 1$.

So it would be a good idea to compute $\pi_1(\text{SO}(3))$. It turns out that $\pi_1(\text{SO}(3)) \simeq \mathbb{Z}/2\mathbb{Z}$. The standard way to visualize this is the plate trick or belt trick, but this doesn’t work very well for me, so here is another way to visualize what’s going on. (I asked a question about this on math.SE and got a very nice answer, but about a slightly different visualization which I’ll describe later.) I became interested in using the visual part of my brain to do mathematics after reading some of Thurston’s excellent answers on MO.

$\text{SO}(3)$ acts transitively on the unit sphere $S^2$ (the superscript denotes its intrinsic dimension, which is why it doesn’t agree with the dimension of the ambient space), and the stabilizer of a point is isomorphic to $\text{SO}(2)$. It follows that $\text{SO}(3)$ acts transitively with trivial stabilizers on the unit tangent bundle $\text{UT}(S^2)$ of $S^2$. Concretely, this is the space of pairs of a point on $S^2$ and a unit tangent vector based at that point. This space is a principal homogeneous space for $\text{SO}(3)$ (an $\text{SO}(3)$-torsor), and in particular is homeomorphic to it, so we can study the fundamental group of $\text{SO}(3)$ by studying paths in $\text{UT}(S^2)$.

$\text{UT}(S^2)$ has a nice visualization as the configuration space of a small tank rolling around on the sphere and pointing its turret in various directions. Since $S^2$ is simply connected (there is a nice visualization of this at the Wikipedia article), any path in $\text{UT}(S^2)$ from a fixed basepoint $b$ to itself (visualized as the journey of the tank around the sphere which ends where it begins, and with the turret pointing in the same direction) is homotopic to a path in which the tank doesn’t move at all, so only its turret moves around in the circle $S^1$. (That is, every path in $\text{SO}(3)$ from the identity to itself is homotopic to a path which consists only of rotations about a fixed axis.) In other words, there is a surjective homomorphism

$\pi_1(S^1) \to \pi_1(\text{UT}(S^2))$.

We know that $\pi_1(S^1) \simeq \mathbb{Z}$ (the winding number), so to determine $\pi_1(\text{UT}(S^2))$ it suffices to determine the smallest positive integer $n$ such that $n$ rotations of a turret is homotopic to the trivial path (if it exists).

So far everything I’ve said has been more or less rigorous, but now I’ll have to turn to the visual explanation. (The results of the next section will give an independent verification of the correct value of $n$.) Consider the path $p$ corresponding to the turret going around twice (that is, twice the image of a generator of $\pi_1(S^1)$). This path is homotopic to a path in which the tank completes a small circle clockwise to the north, keeping the turret pointed inside the circle, then completes a small circle counterclockwise to the south, keeping the turret pointed outside the circle. In $\text{SO}(3)$, this path corresponds to a full rotation about one axis, then a full rotation about another axis close to it. From the perspective of the driver of the tank, looking forward, the turret points constantly to the right.

Now deform the path of the turret so that, while it’s transitioning from the first circle to the second, it is pointing forward most of the time. In $\text{SO}(3)$, this roughly corresponds to ending the first rotation earlier and earlier and starting the second rotation partway through. From the perspective of the tank, the turret points to the right for part of the time, then forward part of the time, then right the rest of the time.

Finally, further deform the path of the turret so that it is always pointing, say, east. I have to admit I have trouble figuring out exactly what this does in $\text{SO}(3)$. From the perspective of the tank, the turret points to the right, then to the left, then to the right again. But now that the turret is always pointing in the same direction (relative to an observer looking down from above), we can deform the path of the tank to its starting point, and now the result is the trivial path in $\text{UT}(S^2)$!

It follows that $n | 2$, so $\pi_1(\text{SO}(3))$ is either trivial or $\mathbb{Z}/2\mathbb{Z}$. In fact it is the latter, but this visualization is not well-suited to showing this. There is another one which is (because it also allows a visualization of the universal cover); we’ll get to it in the next post.

Finding the universal cover

$\text{Spin}(3)$ is a compact Lie group with an almost faithful transitive action on the sphere $S^2$. What sort of group might this be? There are a few ways to answer this question, and the one we will use is the following: $S^2$ can be thought of as the Riemann sphere, or abstractly as the complex projective line $\mathbb{P}^1(\mathbb{C})$ parameterizing lines through the origin in $\mathbb{C}^2$. The general linear group $\text{GL}_2(\mathbb{C})$ acts transitively on such lines, although not faithfully, and the appropriate quotient of it gives the group of conformal automorphisms of the Riemann sphere. Since rotations are conformal automorphisms, this is a natural group to look at.

It turns out that every compact subgroup of $\text{GL}_2(\mathbb{C})$ is conjugate to a subgroup of the unitary group $\text{U}(2)$ of all linear transformations preserving an inner product on $\mathbb{C}^2$ (hence $\text{U}(2)$ is its maximal compact subgroup). This is because one can average any inner product to a $G$-invariant inner product for compact $G$, but we don’t need to prove this; this is merely by way of motivation. So if we are looking for compact groups acting on the Riemann sphere we should look inside $\text{U}(2)$.

However, the action of $\text{U}(2)$ on $\mathbb{P}^1(\mathbb{C})$ is far from faithful; it factors, of course, through the projective unitary group. In particular, the action ignores scalar multiples of the identity. Any element of $\text{U}(2)$ is the product of a scalar multiple of the identity and an element of the special unitary group $\text{SU}(2)$, the subgroup of elements with determinant $1$, so it follows that we can restrict our attention to $\text{SU}(2)$.

Like the case of the groups $\text{SO}(n)$, the classification of elements of $\text{SU}(n)$ is also straightforward. Again, it follows by the spectral theorem that every element of $\text{SU}(n)$ has an orthonormal basis of eigenvectors, and because the inner product must be preserved, the corresponding eigenvalues all lie on the unit circle and have product $1$. In the special case of $\text{SU}(2)$, this implies that the two eigenvalues are complex conjugates, hence every element of $\text{SU}(2)$ is conjugate to an element of the form

$\displaystyle \left[ \begin{array}{cc} e^{i \theta} & 0 \\ 0 & e^{-i \theta} \end{array} \right]$.

In particular, every element of $\text{SU}(2)$ is path-connected to the identity. (These are the maximal tori in $\text{SU}(2)$ – note the similarity to the maximal tori in $\text{SO}(3)$.)

So here is what we now want to know: which elements of $\text{SU}(2)$ act by rotation on the Riemann sphere, do we get all rotations, and what is the kernel of the corresponding morphism?

An important comment is in order. In order to give coordinate-independent meaning to the statement “an element of $\text{SU}(2)$ acts by rotation on the Riemann sphere” we need to specify how the inner product on $\mathbb{C}^2$ induces a Riemannian metric on the sphere. There is a way to do this called the Fubini-Study metric, but as I learned on on math.SE it is not trivial to show that this metric agrees with the round metric on the sphere (and this is essentially equivalent to what we want to prove).

Instead, we will not work invariantly and we will just work with a specific stereographic projection, the one sending the unit circle in $\mathbb{C}$ to the equator on the sphere:

If $\mathbb{C}^2$ is given the standard inner product, this projection has the following important property: associated to any point $(u : v) \in \mathbb{P}^1(\mathbb{C})$ is an orthogonal complement point $(\bar{v} : -\bar{u})$. In local coordinates, where we identify a point $(u : v)$ with either a point $(z : 1) \in \mathbb{C}$ or the point at infinity, the orthogonal complement of $z$ is given by $- \frac{1}{\bar{z}}$, which on the plane is nothing other than inversion with respect to the unit circle. And the projection we chose turns inversion into antipode – that is, it maps points which are inverse with respect to each other on $\mathbb{C}$ to points which are antipodal on the Riemann sphere.

This is important because any element of $\text{SU}(2)$, acting by fractional linear transformations $z \mapsto \frac{az + b}{cz + d}$ in local coordinates, preserves orthogonal complements (by definition), hence preserves antipodes, so we are really talking about a well-defined subgroup of the group of conformal automorphisms of an actual sphere. This condition already implies that any element of $\text{U}(2)$ has antipodal fixed points (corresponding to its eigenvectors) which can be neither attractive nor repelling (since its eigenvalues have absolute value $1$), which I believe is already enough to show that they must act as rotations. But I can’t quite prove this directly, so we’ll have to do an additional computation. Expanding in local coordinates about a fixed point, the classification of maximal tori, we see that any element of $\text{SU}(2)$ acts locally as

$\displaystyle z \mapsto \frac{e^{i \theta} z}{e^{-i \theta}} = e^{2 i \theta} z$

hence (by inspecting the stereographic projection) acts as rotation about the axis determined by the fixed points with an angle $2 \theta$. So there is a morphism $\text{SU}(2) \to \text{SO}(3)$ as desired, and it is clearly surjective. It induces $2$-fold covers when restricted to maximal tori (rotations about a fixed axis), so the only element of the kernel is the element $-I \in \text{SU}(2)$ (corresponding to $\theta = \pi$). Hence $\text{SU}(2) / \{ \pm I \} \simeq \text{SO}(3)$.

It remains to verify that $\text{SU}(2)$ is simply connected, and then we will have found our universal cover (and verified that $\pi_1(\text{SO}(3)) \simeq \mathbb{Z}/2\mathbb{Z}$). This is straightforward. With respect to the standard basis and inner product on $\mathbb{C}^2$ the elements of $\text{SU}(2)$ are represented by self-adjoint (Hermitian) matrices of determinant $1$, which are precisely those of the form

$\displaystyle \left[ \begin{array}{cc} a & b \\ - \overline{b} & \overline{a} \end{array} \right]$

where $a, b \in \mathbb{C}$ satisfy $|a|^2 + |b|^2 = 1$. Writing $a = a_x + a_y i, b = b_x + b_y i$, this gives $a_x^2 + a_y^2 + b_x^2 + b_y^2 = 1$, hence $\text{SU}(2)$ is diffeomorphic to the $3$-sphere $S^3$, and all spheres are simply connected.

So $\text{SU}(2) \simeq \text{Spin}(3)$ is the universal cover we were looking for. In particular, the ordinary representations of $\text{SU}(2)$ give the projective representations of $\text{SO}(3)$, so now we see that we need to study the ordinary representations of $\text{SU}(2)$.

### 6 Responses

1. Nice post, I enjoyed reading it very much :) There wasn’t much QM but for a physicist it is sometimes enjoyable to watch the math that makes stuff tick (and be sure that all of this is omitted in almost all physics lectures).

I would like to remark that to find the fundamental group of SO(3) one might also take the approach of studying the Spin(3) directly, i.e. as a group sitting inside a Clifford Algebra (this is also very useful from representation-theoretic point of view). Or if one wants to omit the CA stuff, one can just note that Spin(3) is isomorphic to unit quaternions and start from there. Ultimately all of these approaches are of course connected but I suppose some of the proofs can be shorter and/or cleaner.

• At some point I will try to talk about spin and angular momentum, although I still have to digest it a little. For the special case of SO(3) the Clifford algebra is just the quaternions anyway, which is the next post. But I have to admit I have no intuition for what the meaning of the Clifford algebra is in general. It seems to be geometric and physical in nature.

2. […] Comments « SO(3) and SU(2) […]

3. on March 7, 2011 at 9:44 am | Reply Dan Petersen

Here is a slightly simpler proof that a Möbius transformation that preserves antipodes is a rotation. It will have two fixed points as you say. By conjugating with an element of SO(3) we can assume that these are 0 and the point at infinity. Then the Möbius transformation is given by multiplication by a nonzero scalar a. Now if |a| is not 1, then any two points on the equator (unit circle) will both be mapped into the northern or southern hemisphere, contradicting that the map preserves antipodes.

4. Dear Qiachou Yuan: love your site and also your keen and pithy insights on Math Overflow and Math SE. Very simple question: what did you use to draw that beautiful stereographic projection. I’ve seen the same software used a great deal on Wikipedia math entries (perhaps by you).

• I didn’t draw it! I stole it from Wikipedia.